14 PENDULUM DYNAMICS AND LINEARIZATION
14
35
Pendulum Dynamics and Linearization
Consider a single-link arm, with length l and all the mass m concentrated at the end. A
motor at the fixed pivot point supplies a controllable torque τ . As drawn, a positive torque
drives the arm counter-clockwise, so as to drive the arm angle θ positive. The arm is free to
swing around the full 360 degrees; gravity pulls the arm downward. There is no damping.
m
g
l
W
T
1. Derive and state the equation of motion for this system.
Solution: It is a basic rotary moment of inertia with a gravity effect and input torque.
We will get
ml2 θ¨ = τ − mgl cos θ.
2. For each of the linearization angles [-90,0,90] degrees, answer the following:
(a) What is the static torque needed to support the arm in this configuration?
Solution: For the three angles given, the static torque is the amount needed to
just balance the gravity torque. These values are [0, mgl, 0]; note that the upward
and downward configurations do not take any static torque to maintain.
(b) Assuming that the static torque is continuously applied, but that there is no dy­
namic torque, what is the equation of motion for small deviations from the static
angle? We are looking for a differential equation in the perturbation of θ, say θ̃.
Solution: The key here is to write the angles and the torques as being comprised
¯ tau)
˜ τ̃ ):
¯ and a dynamic part (θ,
of a static (θ,
ml2
�
d �
θ̄ + θ̃ = τ̄ + τ̃ − mgl cos(θ̄ + θ̃) + d
dt
d
ml2 θ̃ = τ̄ − mgl cos θ̄ cos θ̃ − mgl sin θ̄ sin θ̃ + d.
dt
Along with a trigonometric identity, we used the facts that dynamic torque τ̃ is
zero, and that static angle θ̄ has zero second derivative. We can see that cos θ̃ 14 PENDULUM DYNAMICS AND LINEARIZATION
36
linearizes to simply one, and so the first two terms on the right-hand side in the
second line give us precisely the static torques of part (i). The term d represents a
torque perturbation. We are left with the following linear equations for the three
static angles [-90, 0, 90] degrees, respectively:
mgl¨θ̃ = −mglθ̃ + d
mglθ̃¨ = d
¨
mglθ̃ = mglθ̃ + d.
(c) In words and in an annotated graph, show what is the response of the equilibrium
system to a (small) torque impulse. Be sure to take into account the fact that
the system could respond differently to impulses of opposite directions.
Solution: The first differential equation above is that of a second-order, undamped
oscillator. A perturbation to the equilibrium will lead to oscillations of constant
size. The second equation above is a ”double integrator,” like a mass on a slid­
ing contact; an impulsive d will set the arm spinning at constant speed, with no
damping or restoring force. Of course, as the arm travels our linearizing assump­
tions (namely that θ̃ is small) fail, and we see some other interesting behavior.
If the perturbation is upward, the static torque will be larger than the gravity
torque, and the arm will accelerate in the positive direction, up and over the top!
If the perturbation is negative, we can again expect an acceleration in the positive
direction for the same reasons, again going up and over. It doesn’t get stuck at
zero degrees because there is no damping and inertia will carry it through. Fi­
nally, the upward static configuration is just plain unstable because it is as if we
have a negative spring term - any perturbation will cause the arm to fall over.
T
t
T = -90 deg
T = 0 deg
T = +90 deg
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2.017J Design of Electromechanical Robotic Systems
Fall 2009
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